Lighting a match results in light and heat. Since energy was not created by the match, what transformation took place?

Answer:

188.7 m

Explanation:

height of bridge above water (h) = 393 m

mass of bungee jumper (m) = 150 kg

length of cord (L) = 78 m

acceleration due to gravity (g) = 9.8 m/s  

initial energy = mgh = 150 x 9.8 x 393 = 577,710 J

since the jumper barely touches the water, the maximum extension of the cord (x) = 393 - 78 = 315 m

from the conservation of energy mgh =

therefore

577,710 =

k = 11.64 N/m

from Hooke's law, force (f) = kx' ⇒ mg = kx'

where x' is the extension of the cord when it comes to rest

150 x 9.8 = 11.64 × x'

x' = 126.3 m

the final height at which the cord comes to a rest = height of the bridge - length of the cord - extension of the cord when it comes to rest

the final height at which the cord comes to a rest = 393  - 78 - 126.3 = 188.7 m

Answer:

Option (2): Sixteen times the first cube

Explanation:

When the linear dimensions of a solid are multiplied by ' R ', the surface area of all or any part of it increases by R² , and the volume of all or any part of it increases by R³ .

If the sides of the second cube are 4 times the sides of the first one, then the second cube has (4²) = sixteen times the surface area of the first one (2), and it has (4³) = 64 times the volume of the first one.

Answer:

4.1. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. In other words, when one object exerts a force on another object, the second object exerts an equal force in the opposite direction on the first object.

4.2. Here's a labeled free-body diagram for Block A:

```

T (tension in the string)

↑

│

│

│

│

│

F (applied force)

──→ (direction of motion)

```

In this diagram, "T" represents the tension in the string, and "F" represents the applied force at an angle of 30° to the horizontal. The arrow indicates the direction of motion.

4.3. To find the frictional force acting on block A as it accelerates, we can use Newton's Second Law:

\[F_{\text{net, A}} = m_A \cdot a\]

Where:

- \(F_{\text{net, A}}\) is the net force acting on block A.

- \(m_A\) is the mass of block A (given as 15 kg).

- \(a\) is the acceleration (given as 2.08 m/s²).

Rearranging the equation to solve for \(F_{\text{net, A}}\):

\[F_{\text{net, A}} = 15 kg \cdot 2.08 m/s² = 31.2 N\]

Now, we need to consider the frictional force, which opposes the motion and acts in the direction opposite to the applied force. So, the frictional force is 31.2 N in the opposite direction of motion, making it:

Frictional force on block A = -31.2 N

However, since you want it in magnitude, it's 31.2 N.

4.4. To calculate the mass of block B, we can use the fact that block A and block B are connected by a string, so they experience the same acceleration. Therefore, we can use the following equation:

\[F_{\text{net, B}} = m_B \cdot a\]

Where:

- \(F_{\text{net, B}}\) is the net force acting on block B, which is the tension in the string.

- \(m_B\) is the mass of block B (unknown).

- \(a\) is the acceleration (given as 2.08 m/s²).

We already calculated that the tension in the string is 31.2 N. Plugging in the values:

\[31.2 N = m_B \cdot 2.08 m/s²\]

Now, solving for \(m_B\):

\[m_B = \frac{31.2 N}{2.08 m/s²} \approx 15 kg\]

So, the mass of block B is approximately 15 kg.